volatile keyword seems to be useless?

Question

import java.util.concurrent.CountDownLatch;
import java.util.concurrent.atomic.AtomicInteger;

public class Main implements Runnable {

   private final CountDownLatch cdl1 = new CountDownLatch(NUM_THREADS);
   private volatile int bar = 0;
   private AtomicInteger count = new AtomicInteger(0);

   private static final int NUM_THREADS = 25;

   public static void main(String[] args) {
      Main main = new Main();
      for(int i = 0; i < NUM_THREADS; i++)
         new Thread(main).start();
   }

   public void run() {
      int i = count.incrementAndGet();
      cdl1.countDown();
      try {
         cdl1.await();
      } catch (InterruptedException e1) {
         e1.printStackTrace();
      }
      bar = i;
      if(bar != i)
         System.out.println("Bar not equal to i");
      else
         System.out.println("Bar equal to i");
   }

}

Each Thread enters the run method and acquires a unique, thread confined, int variable i by getting a value from the AtomicInteger called count. Each Thread then awaits the CountDownLatch called cdl1 (when the last Thread reaches the latch, all Threads are released). When the latch is released each thread attempts to assign their confined i value to the shared, volatile, int called bar.

I would expect every Thread except one to print out "Bar not equal to i", but every Thread prints "Bar equal to i". Eh, wtf does volatile actually do if not this?

It is a deliberate intention that each Thread attempts to set the value of bar at exactly the same time.

EDIT:

In light of the answer, changed code to this:

...
bar = i;
try {
    Thread.sleep(0);
} catch(InterruptedException e) {
    e.printStackTrace();
}
...

To ensure that a little time is wasted between the set and read of the variable.

Now the print is 50/50 on same/different value for Bar.

Answer 1

The JVM decides when the threads run, not you. If it felt like holding one of the ones whose latch just released for another 10ms, just because, it can do that. After the latch releases, they still have to wait for their turn to execute. Unless you're running it on a 25 core computer, they're not all assigning bar at anywhere near 'the same time' down inside the machine. Since all you're doing is a couple of primitive operations, it's extremely unlikely that one of them won't finish inside its time slice before the next one gets released!

Answer 2

It's not. You're misusing it. There is a great article here by Herb Sutter that explains it in more detail.

The basic idea is that volatile makes variables unoptimisable. It does not make them thread safe.

Answer 3

To answer the 'WTF does volatile actually do?':

volatile is all about visibility. In Java's thread model, if a thread A writes into a regular shared field, there is no guarantee that a thread B will ever see the value written by A, unless the threads are synchronized somehow. volatile is one of the synchronization mechanisms.

Unlike non-volatile fields, when thread A writes into a volatile field and thread B later reads it, B is guaranteed to see the new value and not an older version.

(Actually volatile does even more - thread B will not only see the new value of the field, but everything else written by A before it set the volatile variable as well. It established a happened-before relationship).

Answer 4

What you should do is replace your instance of volatile int with AtomicInteger. See here.

Answer 5

I think you meant to write this:

import java.util.concurrent.CountDownLatch;
import java.util.concurrent.atomic.AtomicInteger;

public class Main implements Runnable {

   private final CountDownLatch cdl1 = new CountDownLatch(NUM_THREADS);
   private volatile int bar = 0;
   private AtomicInteger count = new AtomicInteger(0);

   private static final int NUM_THREADS = 25;

   public static void main(String[] args) {
      Main main = new Main();
      for(int i = 0; i < NUM_THREADS; i++)
         new Thread(main).start();
   }

   public void run() {
      int i = count.incrementAndGet();
      bar = i;
      cdl1.countDown();
      try {
         cdl1.await();
      } catch (InterruptedException e1) {
         e1.printStackTrace();
      }
      if(bar != i)
         System.out.println("Bar not equal to i");
      else
         System.out.println("Bar equal to i");
   }

}

Which prints "Bar not equal to i" like you expected.